2002 AMC 12B Problems/Problem 6
- The following problem is from both the 2002 AMC 12B #6 and 2002 AMC 10B #10, so both problems redirect to this page.
Contents
Problem
Suppose that and are nonzero real numbers, and that the equation has solutions and . Then the pair is
Solutions
Solution 1
Since , it follows by comparing coefficients that and that . Since is nonzero, , and . Thus .
Solution 2
Another method is to use Vieta's formulas. The sum of the solutions to this polynomial is equal to the opposite of the coefficient, since the leading coefficient is 1; in other words, and the product of the solutions is equal to the constant term (i.e, ). Since is nonzero, it follows that and therefore (from the first equation), . Hence,
Solution 3
Since and are solutions to the given equation, we can write the two equations and From the first equation, we get that and substituting this in our second equation, we get that and solving this gives us the solutions and We discard the first two solutions, as the first one doesnt show up in the answer choices and we are given that is nonzero. The only valid solution is then which gives us and
Solution 4 (Using the Answer Choices)
Note that for roots and , . This implies that is , and there is only one answer choice with in the position for , hence,
Video Solution
https://youtu.be/5QdPQ3__a7I?t=498
~ pi_is_3.14
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.